Why $\cos^2 x-\sin^2 x = \cos 2x\;?$ I was hoping someone could explain how $\cos^2 x-\sin^2 x = \cos 2x$
After using the product rule to differentiate $\sin x \cdot \cos x$ I get the answer $\cos^2 x - \sin ^2 x$ I've come across this problem twice now and each time I've gotten the same wrong so I'm hoping someone can point out what I'm missing.
 A: $$\cos(2x) \equiv \underbrace{\cos(\color{green}{x}+\color{red}x) \equiv \cos(\color{green}x)\cos(\color{red}x)-\sin(\color{green}x)\sin(\color{red}x)}_{\text{addition identity for} \ \  \cos(\color{green}\alpha+\color{red}\beta)} \equiv \cos^2(x)-\sin^2(x).$$
A: $\cos^2 (x)-\sin^2(x)=\cos(x)\cos(x)-\sin(x)\sin(x)= \cos(x+x)=\cos(2x)$
A: Algebraic proof. $$\cos(2x)=\cos(x+x)=\cos x \cos x - \sin x \sin x = \cos^2 x - \sin^2 x$$
Using de Moivre's formula. $$\cos(2x) + i \sin(2x) = (\cos x + i \sin x)^2 = \\ \cos^2 x + i^2 \sin^2 x + 2i \cos x \sin x = \cos^2 x - \sin^2 x + 2i \cos x \sin x .$$ 
From here we get $\cos(2x) = \cos^2 x - \sin^2 x.$
You can find more hints at ProofWiki.
A: It can be proved without using trigonometric identity $\cos2x=\cos^2x-\sin^2x$.
$$\cos^2x-\sin^2x=\frac{1+\cos2x}{2}-\frac{1-\cos2x}{2}=\frac{1+\cos2x-1+\cos2x}{2}=\frac{2\cos2x}{2}=\cos2x$$
A: let $OPN$ be an isosceles triangle with $OP=ON=1$ and $O\hat PN = x$ 
let $H$ be the midpoint of $NP$ and let $M$ be the foot of the perpendicular from $P$ to the continuation of  $NO$ 
since the right-angled triangles $NHO$ and $NMP$ are similar we have:
$$
\frac{NP}{NM} = \frac{ON}{HN}
$$
now, using simple geometry and elementary trig on right-angled triangles we have
$$
\begin{align}
HN &= \cos x \\
ON &= 1\\
NP &= 2\cos x\\
NM &=1+\cos 2x
\end{align}
$$
thus
$$
\frac{2\cos x}{1 +\cos 2x} = \frac1{\cos x}
$$
or 
$$
\cos 2x = 2 \cos^2 x - 1
$$
but for all $x$,
$$
1 = \cos^2x + \sin^2 x
$$
giving:
$$
\cos 2x = 2 \cos^2 x - ( \cos^2x + \sin^2 x )
$$
and the required result immediately follows
A: the same diagram also gives an easy demonstration of the fact that 
$$
\sin 2x = 2 \sin x \cos x
$$
as @Sawarnak hinted, with the help of this result, you may apply your original idea to use calculus for an easy derivation, since differentiation gives
$$
2 \cos 2x = 2(\cos^2 x - \sin^2 x)
$$
it is not a bad idea to familiarize yourself with several different 'proofs' of such fundamental identities. this helps to clarify the different assumptions required, and leads to insights about the relationship between different mathematical idioms.
for example if $z$ is a complex number we have:
$$
\mathfrak{Re}(z^2) =  \mathfrak{Re}(z)^2 - \mathfrak{Im}(z)^2
$$
and this leads to a slightly different demonstration resting on the identity:
$$
\mathfrak{arg}(z^2) = 2 \mathfrak{arg}(z)
$$
